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sukhopar [10]
3 years ago
10

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

table (initially at rest) begins to rotate with its rate of rotation constantly increasing. 1)What is the first event that will occur
Physics
1 answer:
inn [45]3 years ago
6 0

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the  frictional force equals F_{Friction} = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore   ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

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Svetllana [295]

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced  by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

where;

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The work done by the force is determined from the dot product of the force and the displacement of the object.

F = (2xyi + 3yj).(a + b)

W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.

How much work does the force do?

Learn more about work done here: brainly.com/question/8119756

5 0
2 years ago
In his novel From the Earth to the Moon (1866), Jules Verne describes a spaceship that is blasted out of 12,000 yards/s. the Col
saw5 [17]

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration (a), measured in meters per square second, is estimated by this kinematic formula:

a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s } (1)

Where:

\Delta s - Travelled distance, measured in meters.

v_{o}, v - Initial and final speeds of the spaceship, measured in meters.

If we know that v_{o} = 0\,\frac{m}{s}, v = 10968\,\frac{m}{s} and \Delta s = 212.8\,m, then the acceleration experimented by the spaceship is:

a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}

a = 282652.782\,\frac{m}{s^{2}}

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

5 0
3 years ago
How much heat is needed to raise the temperature of 8g of water by 20oC?
Elden [556K]
The following information are given in the question:
Mass, M = 8 g
Temperature, T = 20 degree Celsius
Specific heat of water [this value is a constant] C  = 1 c/gc
Heat, Q = ?
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7 0
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Hope it helped! :>

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N s

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<span>kg m/s</span>

4 0
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